How do I plot a graph of y=x^3-9x?
To begin with we see that y=x3-9x is a positive cubic function, and therefore we know its general shape.The first step is to find the x interecepts. We do this by setting y=0 and solving the resulting equation. We can take out a factor of x to give x(x2-9)=0. We notice this is a difference of two squares so can factorise it further to give x(x+3)(x-3)=0. Therefore the curve crosses the x axis at (-3,0), (0,0) and (3,0).The second step is to find the y intercepts. Similarly, we do this by setting x=0 and solving the resulting equation. Therefore the curve crosses the y axis at a single point, the orgin (0,0), which we already knew from the previous step.Since we know the general shape we could now plot the curve, however it is useful to know how to find turning points as well. We find the turning points by differentiating y and setting this equal to zero. We see that dy/dx=3x2-9. Therefore we have 3x2-9=0 which has solutions x=sqrt(3) and x=-sqrt(3). We can easily plug in numbers and hence plot turning points at (-1.73.10.39) and (1.73,-10.39).To find whether these turning points are maximum or minimum we can work out the second derivative of y . If it is positive we have a minimum, if it is negative we have a maximum. Since dy2/dx2=6x, if we plug in x=-sqrt(3) we have a negative second derivative and therefore a maximum as expected. Similarly, at the second turning point x=sqrt(3), we obtain a positive second derivative and therfore a minimum as expected.
